Dan W.

Dan W.

£20 - £22 /hr

Chemical Physics with Industrial Experience (Integrated Masters) - Bristol University

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About me

I am an enthusiastic and engaging student at the University of Bristol. I study Chemical Physics, meaning I have lectures based around both chemistry and physics. I like to think of myself as approachable and friendly, but also know when to knuckle down and take care of a task at hand.


I studied chemistry, maths, physics and biology at A level meaning I have a broad knowledge of each of the sciences and what they require. I achieved an A* in chemistry and A's in my other A levels and always revel in the opportunity to transfer my knowledge to other students. I have helped mentor GCSE maths classes whilst studying for my A levels and frequently helped out my fellow students throughout sixth form.


I find that the best way to understand a problem is to break it down into small steps. Whether it be physics, maths or chemistry, the best way to approach a problem is to understand what you need to work out, what information you are given and then using your knowledge and study, the best pathway to get there.


I hope to give you the best possible learning experience for your given subject!

I am an enthusiastic and engaging student at the University of Bristol. I study Chemical Physics, meaning I have lectures based around both chemistry and physics. I like to think of myself as approachable and friendly, but also know when to knuckle down and take care of a task at hand.


I studied chemistry, maths, physics and biology at A level meaning I have a broad knowledge of each of the sciences and what they require. I achieved an A* in chemistry and A's in my other A levels and always revel in the opportunity to transfer my knowledge to other students. I have helped mentor GCSE maths classes whilst studying for my A levels and frequently helped out my fellow students throughout sixth form.


I find that the best way to understand a problem is to break it down into small steps. Whether it be physics, maths or chemistry, the best way to approach a problem is to understand what you need to work out, what information you are given and then using your knowledge and study, the best pathway to get there.


I hope to give you the best possible learning experience for your given subject!

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About my sessions

For my Online Lessons, I can provide a range of support.


Depending on what the student wants, I can either provide planned teaching sessions on a specific topic of choice (given the student mentions what topics they need support on) or I can provide questions and answers on a specific topic to check that they have understood the concepts or a mixture of both.


I can also provide examination technique support, how to approach certain types of questions and calmly work through them.


Essentially I shall provide whatever support a student requires, I can tailor my lessons to suit their needs and can be as flexible or as stringent as required.

For my Online Lessons, I can provide a range of support.


Depending on what the student wants, I can either provide planned teaching sessions on a specific topic of choice (given the student mentions what topics they need support on) or I can provide questions and answers on a specific topic to check that they have understood the concepts or a mixture of both.


I can also provide examination technique support, how to approach certain types of questions and calmly work through them.


Essentially I shall provide whatever support a student requires, I can tailor my lessons to suit their needs and can be as flexible or as stringent as required.

Show more

Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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21/11/2018

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Qualifications

SubjectQualificationGrade
ChemistryA-level (A2)A*
PhysicsA-level (A2)A
MathsA-level (A2)A
BiologyA-level (A2)A
Certificate in Financial StudiesA-level (AS)A

General Availability

MonTueWedThuFriSatSun
Pre 12pm
12 - 5pm
After 5pm

Pre 12pm

12 - 5pm

After 5pm
Mon
Tue
Wed
Thu
Fri
Sat
Sun

Subjects offered

SubjectQualificationPrices
ChemistryA Level£22 /hr
MathsA Level£22 /hr
PhysicsA Level£22 /hr
ChemistryGCSE£20 /hr
MathsGCSE£20 /hr
PhysicsGCSE£20 /hr

Questions Dan has answered

Given the reaction: H2SO4 + NaOH --> ? + H2O. (a). Work out the salt produced (?) and (b). calculate the pH of the remaining solution when 1.2 g of NaOH and 4.41 g of H2SO4 were added in a 500 ml solution. Of the unreacted H2SO4 95% dissociated.

(a). Firstly, to work out the product, take 2Na+ + SO2- --> 2Na2SO4.(b). To calculate the pH, the equation: pH = -log10([H+]) must be used. Therefore we need to find the concentration of hydrogen ions in the resulting solution, [H+].Step 1: Work out the molar quantities of the reactants.Looking in a periodic table, the molar mass of S = 32 g mol-1, H = 1 g mol-1, Na = 23 g mol-1, O = 16 g mol-1. Therefore:H2SO4 = 2(1) + 32 + 4(16) = 98 g mol-1. NaOH = 23 + 16 + 1 = 40 g mol-1. To work out the molar quantities look at the units or the equation: moles = mass/Mr where Mr is relative molecular mass.for H2SO4 = 4.41 g / 98 g mol-1 = 0.045 mol, for NaOH = 1.2 g / 40 g mol-1 = 0.030 mol.Step 2: Balance the chemical equation.To do this, count the amount of each atom on each side of the reaction equation then add in the appropriate mole fraction to make sure there is the same amount of each atom on both sides. With the above equation on the LHS: 3 x H, 1 x Na, 5 x O, 1 x S. On the RHS: 2 x H, 2 x Na, 5 x O, 1 x S. Therefore the balanced equation should be:H2SO4 + 2NaOH --> Na2SO4 + 2H2O Counting atoms: LHS: 4 x H, 2 x Na, 6 x O, 1 x S. RHS: 4 x H, 2 x Na, 6 x O, 1 x S therefore balanced.Step 3: Calculate concentration of H+ ions remaining.Since the mole ratio of H2SO4 : NaOH = 1:2, this means that the H2SO4 is in excess as only 0.030 mols/2 = 0.015 mols of H2SO4 has reacted. Therefore to calculate how many moles of H2SO4 are left: 0.045 - 0.015 = 0.030 mols. Since only 95% of the unreacted H2SO4 dissociated: 0.030 mols * 0.95 = 0.0285 mols dissociated. Each mol of H2SO4 donates 2 mols of H+ ions, therefore the mols of H+ ions = 0.0285 mols * 2 = 0.057 mols.Therefore the concentration of H+ ions in a 500 ml = 0.5 dm3 solution, [H+] = 0.057 mols/0.5 dm3 = 0.114 mol dm-3. Step 4: Calculating the pH of the solution.Using the equation for pH shown above: pH = -log10([H+]) we can now work out the pH. pH = -log10(0.114 mol dm-3) = 0.94(a). Firstly, to work out the product, take 2Na+ + SO2- --> 2Na2SO4.(b). To calculate the pH, the equation: pH = -log10([H+]) must be used. Therefore we need to find the concentration of hydrogen ions in the resulting solution, [H+].Step 1: Work out the molar quantities of the reactants.Looking in a periodic table, the molar mass of S = 32 g mol-1, H = 1 g mol-1, Na = 23 g mol-1, O = 16 g mol-1. Therefore:H2SO4 = 2(1) + 32 + 4(16) = 98 g mol-1. NaOH = 23 + 16 + 1 = 40 g mol-1. To work out the molar quantities look at the units or the equation: moles = mass/Mr where Mr is relative molecular mass.for H2SO4 = 4.41 g / 98 g mol-1 = 0.045 mol, for NaOH = 1.2 g / 40 g mol-1 = 0.030 mol.Step 2: Balance the chemical equation.To do this, count the amount of each atom on each side of the reaction equation then add in the appropriate mole fraction to make sure there is the same amount of each atom on both sides. With the above equation on the LHS: 3 x H, 1 x Na, 5 x O, 1 x S. On the RHS: 2 x H, 2 x Na, 5 x O, 1 x S. Therefore the balanced equation should be:H2SO4 + 2NaOH --> Na2SO4 + 2H2O Counting atoms: LHS: 4 x H, 2 x Na, 6 x O, 1 x S. RHS: 4 x H, 2 x Na, 6 x O, 1 x S therefore balanced.Step 3: Calculate concentration of H+ ions remaining.Since the mole ratio of H2SO4 : NaOH = 1:2, this means that the H2SO4 is in excess as only 0.030 mols/2 = 0.015 mols of H2SO4 has reacted. Therefore to calculate how many moles of H2SO4 are left: 0.045 - 0.015 = 0.030 mols. Since only 95% of the unreacted H2SO4 dissociated: 0.030 mols * 0.95 = 0.0285 mols dissociated. Each mol of H2SO4 donates 2 mols of H+ ions, therefore the mols of H+ ions = 0.0285 mols * 2 = 0.057 mols.Therefore the concentration of H+ ions in a 500 ml = 0.5 dm3 solution, [H+] = 0.057 mols/0.5 dm3 = 0.114 mol dm-3. Step 4: Calculating the pH of the solution.Using the equation for pH shown above: pH = -log10([H+]) we can now work out the pH. pH = -log10(0.114 mol dm-3) = 0.94

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3 months ago

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